# 2031. Count Subarrays With More Ones Than Zeros

## Description

You are given a binary array nums containing only the integers 0 and 1. Return the number of subarrays in nums that have more 1's than 0's. Since the answer may be very large, return it modulo 109 + 7.

A subarray is a contiguous sequence of elements within an array.

Example 1:

Input: nums = [0,1,1,0,1]
Output: 9
Explanation:
The subarrays of size 1 that have more ones than zeros are: [1], [1], [1]
The subarrays of size 2 that have more ones than zeros are: [1,1]
The subarrays of size 3 that have more ones than zeros are: [0,1,1], [1,1,0], [1,0,1]
The subarrays of size 4 that have more ones than zeros are: [1,1,0,1]
The subarrays of size 5 that have more ones than zeros are: [0,1,1,0,1]


Example 2:

Input: nums = [0]
Output: 0
Explanation:
No subarrays have more ones than zeros.


Example 3:

Input: nums = [1]
Output: 1
Explanation:
The subarrays of size 1 that have more ones than zeros are: [1]


Constraints:

• 1 <= nums.length <= 105
• 0 <= nums[i] <= 1

## Solutions

Solution 1: Prefix Sum + Binary Indexed Tree

The problem requires us to count the number of subarrays where the count of $1$ is greater than the count of $0$. If we treat $0$ in the array as $-1$, then the problem becomes counting the number of subarrays where the sum of elements is greater than $0$.

To calculate the sum of elements in a subarray, we can use the prefix sum. To count the number of subarrays where the sum of elements is greater than $0$, we can use a binary indexed tree to maintain the occurrence count of each prefix sum. Initially, the occurrence count of the prefix sum $0$ is $1$.

Next, we traverse the array $nums$, use variable $s$ to record the current prefix sum, and use variable $ans$ to record the answer. For each position $i$, we update the prefix sum $s$, then query the occurrence count of the prefix sum in the range $[0, s)$ in the binary indexed tree, add it to $ans$, and then update the occurrence count of $s$ in the binary indexed tree.

Finally, return $ans$.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $nums$.

• class BinaryIndexedTree {
private int n;
private int[] c;

public BinaryIndexedTree(int n) {
this.n = n;
c = new int[n + 1];
}

public void update(int x, int v) {
for (; x <= n; x += x & -x) {
c[x] += v;
}
}

public int query(int x) {
int s = 0;
for (; x > 0; x -= x & -x) {
s += c[x];
}
return s;
}
}

class Solution {
public int subarraysWithMoreZerosThanOnes(int[] nums) {
int n = nums.length;
int base = n + 1;
BinaryIndexedTree tree = new BinaryIndexedTree(n + base);
tree.update(base, 1);
final int mod = (int) 1e9 + 7;
int ans = 0, s = 0;
for (int x : nums) {
s += x == 0 ? -1 : 1;
ans += tree.query(s - 1 + base);
ans %= mod;
tree.update(s + base, 1);
}
return ans;
}
}

• class BinaryIndexedTree {
private:
int n;
vector<int> c;

public:
BinaryIndexedTree(int n)
: n(n)
, c(n + 1, 0) {}

void update(int x, int v) {
for (; x <= n; x += x & -x) {
c[x] += v;
}
}

int query(int x) {
int s = 0;
for (; x > 0; x -= x & -x) {
s += c[x];
}
return s;
}
};

class Solution {
public:
int subarraysWithMoreZerosThanOnes(vector<int>& nums) {
int n = nums.size();
int base = n + 1;
BinaryIndexedTree tree(n + base);
tree.update(base, 1);
const int mod = 1e9 + 7;
int ans = 0, s = 0;
for (int x : nums) {
s += (x == 0) ? -1 : 1;
ans += tree.query(s - 1 + base);
ans %= mod;
tree.update(s + base, 1);
}
return ans;
}
};

• class BinaryIndexedTree:
__slots__ = ["n", "c"]

def __init__(self, n: int):
self.n = n
self.c = [0] * (n + 1)

def update(self, x: int, v: int):
while x <= self.n:
self.c[x] += v
x += x & -x

def query(self, x: int) -> int:
s = 0
while x:
s += self.c[x]
x -= x & -x
return s

class Solution:
def subarraysWithMoreZerosThanOnes(self, nums: List[int]) -> int:
n = len(nums)
base = n + 1
tree = BinaryIndexedTree(n + base)
tree.update(base, 1)
mod = 10**9 + 7
ans = s = 0
for x in nums:
s += x or -1
ans += tree.query(s - 1 + base)
ans %= mod
tree.update(s + base, 1)
return ans


• type BinaryIndexedTree struct {
n int
c []int
}

func newBinaryIndexedTree(n int) *BinaryIndexedTree {
return &BinaryIndexedTree{n: n, c: make([]int, n+1)}
}

func (bit *BinaryIndexedTree) update(x, v int) {
for ; x <= bit.n; x += x & -x {
bit.c[x] += v
}
}

func (bit *BinaryIndexedTree) query(x int) (s int) {
for ; x > 0; x -= x & -x {
s += bit.c[x]
}
return
}

func subarraysWithMoreZerosThanOnes(nums []int) (ans int) {
n := len(nums)
base := n + 1
tree := newBinaryIndexedTree(n + base)
tree.update(base, 1)
const mod = int(1e9) + 7
s := 0
for _, x := range nums {
if x == 0 {
s--
} else {
s++
}
ans += tree.query(s - 1 + base)
ans %= mod
tree.update(s+base, 1)
}
return
}

• class BinaryIndexedTree {
private n: number;
private c: number[];

constructor(n: number) {
this.n = n;
this.c = Array(n + 1).fill(0);
}

update(x: number, v: number): void {
for (; x <= this.n; x += x & -x) {
this.c[x] += v;
}
}

query(x: number): number {
let s = 0;
for (; x > 0; x -= x & -x) {
s += this.c[x];
}
return s;
}
}

function subarraysWithMoreZerosThanOnes(nums: number[]): number {
const n: number = nums.length;
const base: number = n + 1;
const tree: BinaryIndexedTree = new BinaryIndexedTree(n + base);
tree.update(base, 1);
const mod: number = 1e9 + 7;
let ans: number = 0;
let s: number = 0;
for (const x of nums) {
s += x || -1;
ans += tree.query(s - 1 + base);
ans %= mod;
tree.update(s + base, 1);
}
return ans;
}